Question 1(Multiple Choice Worth 1 points)
(Rational Functions' Vertical Asymptotes and Holes MC)

The table gives selected values for a function h(x).


x	2.997	2.998	2.999	3.001	3.002	3.003
h(x)	0.142847	0.142850	0.142854	0.142861	0.142864	0.142867


If h is defined as h of x equals the fraction with numerator binomial x squared minus 9 and denominator trinomial 6 x squared plus 6 x minus 72, which of the following statements is true?
 the limit of h of x as x approaches 3 equals 0.143, and h(x) has a vertical asymptote at x = 3.
 the limit of h of x as x approaches 3 equals 0.143, and h(3) = 0.143.
 the limit of h of x as x approaches 3 from the left equals the limit of h of x as x approaches 3 from the right equals 0.143, and h(3) = 0.143.
 the limit of h of x as x approaches 3 equals 0.143, and h(x) has a hole at x = 3.

Let's analyze the function \( h(x) = \frac{x^2 - 9}{6x^2 + 6x - 72} \) and determine its behavior around \( x = 3 \).

### Step 1: Factor the function
First, factor both the numerator and denominator:
- Numerator: \( x^2 - 9 \) is a difference of squares, so it factors as \( (x - 3)(x + 3) \).
- Denominator: \( 6x^2 + 6x - 72 \) can be factored by first factoring out a 6:
  \[
  6(x^2 + x - 12)
  \]
  Then factor \( x^2 + x - 12 \):
  \[
  x^2 + x - 12 = (x - 3)(x + 4)
  \]
  So the denominator becomes \( 6(x - 3)(x + 4) \).

Thus, the function becomes:
\[
h(x) = \frac{(x - 3)(x + 3)}{6(x - 3)(x + 4)}
\]

### Step 2: Simplify the function
Notice that the factor \( (x - 3) \) appears in both the numerator and denominator. For \( x \neq 3 \), these terms can cancel out:
\[
h(x) = \frac{x + 3}{6(x + 4)}
\]

### Step 3: Behavior at \( x = 3 \)
- **Holes**: Since \( x = 3 \) was canceled out in both the numerator and denominator, the function has a **hole** at \( x = 3 \). The function is undefined at this point.
- **Limit**: To find the limit of \( h(x) \) as \( x \) approaches 3, substitute \( x = 3 \) into the simplified form of the function:
  \[
  \lim_{x \to 3} h(x) = \frac{3 + 3}{6(3 + 4)} = \frac{6}{6 \times 7} = \frac{6}{42} = 0.143
  \]
  So the limit as \( x \to 3 \) is 0.143.

### Step 4: Conclusion
The function has a **hole** at \( x = 3 \), and the limit of \( h(x) \) as \( x \) approaches 3 is 0.143. Therefore, the correct statement is:

- **The limit of \( h(x) \) as \( x \) approaches 3 equals 0.143, and \( h(x) \) has a hole at \( x = 3 \).**

This matches the last option.

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Would you like further clarification on this? Here are some related questions for deeper understanding:

1. How do vertical asymptotes differ from holes in rational functions?
2. Can you explain how to find the limit of a function using cancellation of factors?
3. What happens when a function has a factor that cancels out in both the numerator and the denominator?
4. How do you determine if a rational function has a hole or a vertical asymptote?
5. What does it mean when the limit exists but the function is undefined at a point?

**Tip:** Always check for common factors in the numerator and denominator to simplify rational functions and identify potential holes!

The table gives selected values for a function h(x). 

x	2.997	2.998	2.999	3.001	3.002	3.003
h(x)	0.142847	0.142850	0.142854	0.142861	0.142864	0.142867

If h is defined as h(x) = (x^2 - 9) / (6x^2 + 6x - 72), which of the following statements is true?
- The limit of h(x) as x approaches 3 equals 0.143, and h(x) has a vertical asymptote at x = 3.
- The limit of h(x) as x approaches 3 equals 0.143, and h(3) = 0.143.
- The limit of h(x) as x approaches 3 from the left equals the limit of h(x) as x approaches 3 from the right equals 0.143, and h(3) = 0.143.
- The limit of h(x) as x approaches 3 equals 0.143, and h(x) has a hole at x = 3.

Explore how to analyze the rational function h(x) = (x^2 - 9) / (6x^2 + 6x - 72), including factoring, finding limits, and identifying holes or asymptotes. This detailed explanation helps students understand key concepts like rational functions and limits. Suitable for high school students in Grades 10-12.

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